• What is the Maxwell-Boltzmann distribution?
• For which “substances” does the Maxwell-Boltzmann distribution only apply?
• How is the area under the Maxwell-Boltzmann distribution to be interpreted?
• How does the gas pressure and temperature influence the speed distribution of an ideal gas?
• How can the process of evaporation of liquids be qualitatively explained with the Maxwell-Boltzmann distribution?
• With which speeds can the Maxwell-Boltzmann distribution be characterized?
• Why is temperature generally not a direct measure of the temperature of gases?
• What is the relationship between the temperature and the average kinetic energy of the particles in an ideal gas?
• How can an energy distribution function be derived from the speed distribution function?

## Introduction

As already explained in the article “Temperature and particle motion“, the temperature of a gas is a measure of the kinetic energy of the particles it contains. Even at a constant temperature, however, not all the particles have the same speed. After all, in a gas at the atomic level there are permanent collisions between the particles. Some particles are slowed down by the collision and others are accelerated by it. Thus, particles with different velocities can be found in a gas.

The article “Determination of the speed distribution in a gas” deals with the experimental determination of such a speed distribution in more detail. In this article, however, this speed distribution will be explained and interpreted in more detail and the respective distribution function will be derived mathematically.

Due to the large number of molecules that a gas usually contains, it is possible to make statistical predictions about the speed distribution of the gas particles. The so-called Maxwell-Boltzmann distribution describes exactly this speed distribution for the particles of an ideal gas.

The Maxwell-Boltzmann distribution represents the speed distribution of the particles of an ideal gas!

Such a statistical prediction about the speed distribution is only possible if a sufficient number of molecules are present. This is true in most thermodynamic cases! To get an impression of the large number of particles in an ideal gas, imagine a football filled with air. The number of gas particles in such a football would then correspond approximately to the number of 1-liter water bottles that would theoretically be necessary to completely fill the entire volume of the earth with water!

This example clearly shows that in practice a sufficiently large number of particles in a gas can usually be assumed to make reliable statistical predictions.

## Maxwell-Boltzmann speed distribution

Using statistical methods, physicists James Clerk Maxwell and Ludwig Boltzmann were able to derive the following formula for determining the speed distribution of the molecules contained in an ideal gas. For this reason this function of the relative frequency density $$f(v)$$ is also called Maxwell-Boltzmann distribution function (the derivation of this function is shown in a separate article):

\begin{align}
\label{p}
&\boxed{  f(v) = \left(  \sqrt{\frac{m}{2 \pi k_B T}}  \right)^{3} 4 \pi v^2 \cdot \exp{\left(- \frac{m v^2}{2 k_B T} \right)} } ~~~\text{Maxwell-Boltzmann distribution function} \\[5px]
\end{align}

In equation (\ref{p}), $$m$$ denotes the mass of a particle (not the total mass of the gas!). For a helium particle, the mass is $$m = 6.6465 \cdot 10^{-27} \text{ kg}$$. The constant $$k_B$$ is the so-called Boltzmann constant with a value of $$k_B=1,38065 \cdot 10^{-23} \frac{\text{J}}{\text{kg}}$$. The temperature $$T$$ must be expressed in the unit Kelvin! Note that the distribution of the speeds is independent of the gas pressure!

The distribution of particle speeds for an ideal gas is independent of the gas pressure!

## Interpretation of the Maxwell-Boltzmann distribution function

The Maxwell-Boltzmann distribution describes the abundance with which certain molecular speeds occur in an ideal gas. In principle, however, it is not possible to assign a specific number of particles to a specific speed. Ultimately, you will never find a single particle with an exact velocity down to the “last” decimal place. One can only assign speed ranges to a concrete number of particles, i.e. the number of particles whose speeds lie within a certain interval (e.g. between 1300 $$\frac{\text{m}}{\text{s}}$$ and 1600 $$\frac{\text{m}}{\text{s}}$$).

Therefore one chooses a graph in which the area under the curve between two speeds corresponds to the percentage share of the particles whose speed lies within the speed interval considered! Therefore one does not speak of the frequency but of the frequency density(“frequency related to the speed interval”). As it is not an absolute frequency (i.e. a concrete number of particles) but a percentage, it is referred to as the relative frequency density.

The area under the Maxwell-Boltzmann distribution function corresponds to the percentage of particles whose speeds lie within the speed range considered!

More detailed information on the derivation and interpretation of such a speed distribution can be found in the article “Determination of the speed distribution in a gas“!

If, for example, there is an area of 0.2 in the range between 1300 $$\frac{\text{m}}{\text{s}}$$ and 1600 $$\frac{\text{m}}{\text{s}}$$ below the Maxwell-Boltzmann function (see figure above), this means that 20 % of the particles have a speed within this range. If you randomly select 100 molecules in a gas with such a velocity distribution, the statistical average would be 20 particles with velocities within this interval. The probability of catching a particle whose velocity lies within this interval in a single access to a particle would thus be 20 %.

This example shows that the relative frequency density of the Maxwell-Boltzmann distribution function can also be interpreted as the probability density of encountering a randomly selected particle in a certain speed range!

The relative frequency density can also be understood as probability density, i.e. the probability to encounter a particle in a certain speed range when randomly picking it up!

With this interpretation it also becomes clear that if only a certain speed is considered, the area under the curve becomes zero and therefore the probability of finding a particle with exactly this velocity is also zero, i.e. such a particle does not exist with an exactly given speed!

## Influence of temperature on speed distribution

The effects of ever higher temperatures on the Maxwell-Boltzmann distribution are shown in the figure below. As the temperature rises, the curve maximum shifts to higher and higher velocities. The curves are then stretched in length and squeezed in height. This results in a broader speed distribution for high temperatures, with correspondingly higher speed proportions.

With increasing temperatures, the curve maximum shifts to higher and higher speeds!

However, all curves are basically “open to the right”, i.e. even at very low temperatures, particles with very high speeds can always be found!

Even at such low temperatures, there are gas particles that have very high speeds!

If this statement is transferred qualitatively to liquids, then it can also be used to explain why liquids gradually enter the gas phase, i.e. evaporate, even far below the boiling point. Even at such low temperatures, there are always particles that have a sufficiently high velocity and thus kinetic energy to escape the attraction forces within the liquid phase. They practically break away from the binding forces and thus escape from the liquid. The liquid gradually becomes less until the molecules are completely in the gas phase. Since according to the speed distribution the number of particles that have such a sufficiently high velocity is relatively small, such an evaporation process takes a relatively long time compared to the vaporisation process. Further information can be found in the article “Evaporation of liquids“.

## Characteristic speeds

To characterize the speed distribution, different speeds are introduced, which are explained in more detail in the following.

### Most probable speed

If a particle is randomly picked out of an ideal gas, it is most likely to be in the speed range with the highest proportion. This corresponds to the maximum of the speed distribution and is referred to as the most probable speed $$\hat{v}$$.

Mathematically, this most likely speed can be determined by setting the first derivative of the Maxwell-Boltzmann distribution (\ref{p}) to zero $$\left[\frac{df(v)}{dv}=0\right]$$. The result is the following formula for determining the most probable speed:

\begin{align}
\label{w}
&\boxed{ \hat{v} = \sqrt{\frac{2 k_B T}{m}}  } ~~~\text{most probable speed}\\[5px]
\end{align}

The most probable speed is not only a function of temperature but also depends on the mass of the particles. Thus, the most probable speed cannot be deduced directly from the temperature of a gas. Two different gases (whose particles have different masses) therefore have different most probable speeds despite the same temperature (see figure below). For heavier gas particles, the most probable speeds will be lower than for lighter particles.

### Arithmetic mean speed

Another characteristic speed of the Maxwell-Boltzmann distribution is the arithmetic mean speed $$\overline{v}$$ of a particle (also called average speed or just mean speed).

Compared to the most probable speed, the mean speed will be higher, since the number of particles with a higher speed than the most probable is also higher. This becomes clear when the area to the right and to the left of the maximum is compared. 42.8 % of the particles have a velocity below the most probable and 57.2 % have a velocity above the most probable.

The average speed is obtained by summing up the speeds of the particles and then dividing the sum by the number of particles. Mathematically, this can be calculated by solving the integral $$\int \! v~f(v) \, \mathrm{d}v$$ within the entire speed range from 0 to $$\infty$$. The result is as follows:

\begin{align}
&\overline{v} = \frac{v_1+v_2+v_3+…+v_N}{N} =  \frac{\sum_{i=1}^{N}{v_i}}{N} = \int_0^\infty \! v~f(v) \, \mathrm{d}v \\[5px]
\label{a}
&\boxed{ \overline{v} = \sqrt{\frac{8 k_B T}{\pi m}}  } ~~~\text{mean speed} \\[5px]
\end{align}

At this point, too, it can be seen that the temperature is not a direct measure of the average particle speed. The particles in a gas with four times the particle mass are on average only half as fast, even with the same temperature.

To get an impression of the order of magnitude of the mean speed in gases, nitrogen molecules $$N_2$$ are considered, which make up the majority of the air with 78 %. With a particle mass of 4.65 $$\cdot 10^{-27}$$ kg, the nitrogen molecules have a mean speed of about 470 $$\frac{\text{m}}{\text{s}}$$ at room temperature (273 K). Note that the average speed of the nitrogen molecules in the air is thus greater than the speed of sound! In contrast to sound, a nitrogen molecule does not travel several meters or even kilometers in a certain direction. It will permanently collide with other air particles and constantly change direction. Such a distance without collision is usually only a few nanometers and is called the mean free path!

### Root-mean-square speed

However, neither the most probable speed nor the average speed is relevant for determining the kinetic energy of a molecule, because due to the quadratic influence of velocity on kinetic energy, high speeds have a disproportionately strong effect on kinetic energy. Thus, a speed twice as high does not mean a kinetic energy twice as high but four times as high. Higher speeds influence the average kinetic energy more than lower speeds.

Therefore, the mean speed must not be used to characterize the kinetic energy of a particle. Rather, the arithmetic mean of the squares of the speeds must be taken as a basis. In order to obtain the dimension of a speed, the square root of this mean value must be calculated. This speed is then referred to as the root-mean-square speed $$v_{rms}$$:

\begin{align}
&\boxed{v_{rms} = \sqrt{\overline{v^2}}} \\[5px]
\end{align}

For the calculation of the root-mean-square speed from the Maxwell-Boltzmann function, the square root of the integral $$\int \! v^2 f(v) \, \mathrm{d}v$$ has to be calculated within the range between 0 and $$\infty$$:

\begin{align}
& v_{rms} = \sqrt{\overline{v^2}} = \sqrt{\frac{v_1^2+v_2^2+v_3^2+…+v_N^2}{N}} =  \sqrt{\frac{\sum_{i=1}^{N}{v_i^2}}{N}} = \sqrt{\int_0^\infty \! v^2~f(v) \, \mathrm{d}v }\\[5px]
\label{q}
&\boxed{v_{rms} = \sqrt{\frac{3 k_B T}{m}} } ~~~\text{root-mean-square speed} \\[5px]
\end{align}

It now becomes apparent that the root-mean-square speed is decisive for the average kinetic energy of a particle:

\begin{align}
\overline{W_{kin}} &= \tfrac{W_{kin,1}+W_{kin,2}+W_{kin,3} + … + W_{kin,N}}{N} = {\tfrac{\frac{1}{2}m\cdot v_1^2+\frac{1}{2}m\cdot v_2^2+\frac{1}{2}m\cdot v_3^2+…+\frac{1}{2}m\cdot v_N^2}{N}}  \\[5px]
&= \frac{1}{2}m\cdot \underbrace{{\tfrac{v_1^2+v_2^2+v_3^2+…+v_N^2}{N}}}_{v_{rms}^2} = \frac{1}{2} m \cdot  v_{rms}^2  \\[5px]
\end{align}

\begin{align}
\label{e}
\boxed{\overline{W_{kin}} = \frac{1}{2} m \cdot  v_{rms}^2}  \\[5px]
\end{align}

### Relationship between the different speeds

The average speed of the particles $$\overline{v}$$ is always greater than the most probable speed $$\hat{v}$$, and the root-mean-square speed $$v_{rms}$$ is greater than the average speed. The different speeds are in a constant ratio to each other, independent of temperature or mass. For the ratio of average speed $$\overline{v}$$ to most probable speed $$\hat{v}$$ applies:

\begin{align}
\frac{\overline{v}}{\hat{v}}= \frac{\sqrt{\frac{8 k_B T}{\pi m}} }{
\sqrt{\frac{2 k_B T}{m}} } = \sqrt{\frac{4}{\pi}} = 1,128\\[5px]
\end{align}

For the ratio of root-mean-square speed $$v_{rms}$$ to average speed $$\overline{v}$$ a value of 1.085 is obtained:

\begin{align}
\frac{v_{rms}}{\overline{v}}= \frac{\sqrt{\frac{3 k_B T}{m}}}{\sqrt{\frac{8 k_B T}{\pi m}}} = \sqrt{\frac{3\pi}{8}} = 1,085\\[5px]
\end{align}

The average speed is therefore always 12,8 % higher than the most probable speed and the root-mean-square speed is always 8,5 % higher than the average speed.

## Relationship between temperature and kinetic energy

The different speeds (the most probable speed, the average speed and the root-mean-square speed) depend not only on the temperature but also on the particle mass. The frequently heard statement that the temperature is a measure for the speed of the particles is strictly speaking not correct.

For example, such a statement does not apply if two different types of gas are considered. Even at temperatures of 473 K (200 °C) the particle speeds of argon are more than half lower compared to the particle speeds of helium at 273 K (0 °C), because argon has an atomic mass about 40 times larger than helium.

Rather, the statement about the temperature aims at the average kinetic energy of the particles. This becomes clear when the formula for the root-mean-square speed (\ref{q}) is applied to the equation for the average kinetic energy (\ref{e}):

\begin{align}
&\overline{W_{kin}} = \frac{1}{2} m \cdot v_{rms}^2 = \frac{1}{2} m \cdot \frac{3 k_B T}{m} \\[5px]
\label{wkint}
&\boxed{\overline{W_{kin}}=\frac{3}{2}~k_B~T } \\[5px]
\end{align}

The average kinetic energy of a particle is directly connected to the temperature and independent of the particle mass! Thus the temperature is directly a measure for the average kinetic energy of the gas particles of an ideal gas. This equation is remarkable because it combines a macroscopically measurable quantity (in the form of temperature) with a microscopic quantity (in the form of the kinetic energy of a particle)!

The temperature of an ideal gas is directly a measure of the average kinetic energy of the particles!

Equation (\ref{wkint}) was already derived in the article “Pressure and temperature” by kinematic and statistical considerations. It is no coincidence at this point that the Maxwell-Boltzmann distribution comes to the same conclusion. After all, the derivation of the Maxwell-Boltzmann distribution function is based on the assumption that the mean kinetic energy of a particle is linked to the temperature according to the equation (\ref{wkint})!

## Maxwell-Boltzmann kinetic energy distribution

Since a certain kinetic energy can be assigned to each speed, the speed distribution can also be converted into an energy distribution. Instead of the distribution function $$f(v)$$ for the speed, a distribution function for the kinetic energy $$g(W)$$ is obtained.

As already explained, the (relative) frequency with which a speed exists within the limits between $$v_1$$ and $$v_2$$ results from the integral of the distribution function $$f(v)$$:

\begin{align}
& \text{Frequency} = \int_{v_1}^{v_2} \! f(v) \, \mathrm{d}v   \\[5px]
\end{align}

If the speeds $$v_1$$ and $$v_2$$ are converted into the respective kinetic energies $$W_{1}$$ and $$W_{2}$$, then a distribution function of the energy $$g(W_{kin})$$ within these limits must lead to the same frequency:

\begin{align}
& \text{Frequency} = \int_{W_{1}}^{W_{2}} \! g(W) \, \mathrm{d}W   \\[5px]
\end{align}

If both equations are equated, then the following relationship between the two distribution functions is obtained:

\begin{align}
\int_{W_{1}}^{W_{2}} \! g(W) \, \mathrm{d}W &= \int_{v_1}^{v_2} \! f(v) \, \mathrm{d}v \\[5px]
g(W) \text{d}W &=f(v) ~ \text{d}v \\[5px]
g(W) &= f(v) ~ \frac{\text{d}v}{\text{d}W} \\[5px]
g(W) &= f(v) ~ \dfrac{1}{\frac{\text{d}W}{\text{d}v}} \\[5px]
\end{align}

The expression $$\frac{\text{d}W}{\text{d}v}$$ contained in the denominator corresponds mathematically to the speed derivative of the kinetic energy:

\begin{align}
& \frac{\text{d}W(v)}{\text{d}v} = \frac{\text{d}(\tfrac{1}{2}mv^2)}{\text{d}v} = mv \\[5px]
\end{align}

Thus the distribution function of the kinetic energy $$g(W)$$ can be determined as follows from the distribution function of the speed $$f(v)$$:

\begin{align}
&g(W) = f(v) ~ \dfrac{1}{\frac{\text{d}W}{\text{d}v}} = f(v) ~ \dfrac{1}{mv} \\[5px]
&\boxed{g(W) = f(v) ~ \dfrac{1}{mv} } \\[5px]
\end{align}

If the speed distribution $$f(v)$$ is used at this point and the terms are transformed and summarized in such a way that only the kinetic energies can be found in them, then the following distribution function of the kinetic energy becomes apparent:

\begin{align}
\require{cancel}
g(W) &= \left(  \sqrt{\frac{m}{2 \pi k_B T}}  \right)^{3} 4 \pi v^2 \cdot \exp{\left(- \frac{\color{red}{m v^2}}{\color{red}{2} k_B T} \right)} ~ \dfrac{1}{mv} \\[5px]
& = \left(  \sqrt{\frac{m}{2 \pi k_B T}}  \right)^{3} 4 \pi \frac{v^\bcancel{2}}{m\bcancel{v}} \cdot \exp{\left(- \frac{\color{red}{\tfrac{1}{2}m v^2}}{k_B T} \right)} \\[5px]
& = \sqrt{\left(  \frac{m}{2 \pi k_B T}  \right)^{3}} \cdot 4 \pi \frac{v}{m} \cdot \exp{\left(- \frac{\color{red}{\tfrac{1}{2}m v^2}}{k_B T} \right)} \\[5px]
& = \sqrt{\left( \frac{m}{2 \pi k_B T} \right)^{3}} \cdot \sqrt{\left(4 \pi \frac{v}{m} \right)^2} \cdot \exp{\left(- \frac{\color{red}{\tfrac{1}{2}m v^2}}{k_B T} \right)} \\[5px]
& = \sqrt{\left( \frac{m}{2 \pi k_B T} \right)^{3} \cdot \left(4 \pi \frac{v}{m} \right)^2} \cdot \exp{\left(- \frac{\color{red}{\tfrac{1}{2}m v^2}}{k_B T} \right)} \\[5px]
& = \sqrt{ \frac{m^\bcancel{3}}{8 \pi^3 k_B^3 T^3} \cdot 16 \pi \frac{v^2}{\bcancel{m^2}} } \cdot \exp{\left(- \frac{\color{red}{\tfrac{1}{2}m v^2}}{k_B T} \right)} \\[5px]
& = \sqrt{\frac{4}{\pi^2 k_B^3 T^3} \cdot \color{red}{\tfrac{1}{2}mv^2} } \cdot \exp{\left(- \frac{\color{red}{\tfrac{1}{2}m v^2}}{k_B T} \right)} \\[5px]
& = \frac{2}{\sqrt{\pi}}\sqrt{\left(\frac{1}{k_B T}\right)^3} \cdot \sqrt{\color{red}{\tfrac{1}{2}mv^2}} \cdot \exp{\left(- \frac{\color{red}{\tfrac{1}{2}m v^2}}{k_B T} \right)} \\[5px]
& = \frac{2}{\sqrt{\pi}}\left(\sqrt{\frac{1}{k_B T}}\right)^3 \cdot \sqrt{\color{red}{\tfrac{1}{2}mv^2}} \cdot \exp{\left(- \frac{\color{red}{\tfrac{1}{2}m v^2}}{k_B T} \right)} \\[5px]
\end{align}

The red marked terms correspond to the kinetic energy, so the distribution function can be expressed solely by the kinetic energy:

\begin{align}
&\boxed{g(W) = \frac{2}{\sqrt{\pi}}\left(\sqrt{\frac{1}{k_B T}}\right)^3 \cdot \sqrt{W} \cdot \exp{\left(- \frac{W}{k_B T} \right)}} \\[5px]
\end{align}

The figure below shows the distribution of the kinetic energies for a temperature of 500 K. Again, the area under the curve corresponds to the percentage of particles with kinetic energy within the corresponding limits.

In the article Derivation of the Maxwell-Boltzmann distribution function, the distribution function of the molecular speeds was derived from the barometric formula. It has already been noted that the exponential term $$\exp\left(-\frac{W}{k_BT}\right)$$ describes the frequency or probability with which certain (kinetic) energies are present. This term plays a central role in statistical physics (Boltzmann statistics) and occurs whenever random distributions of energetic states are involved. For example also with Planck’s law of blackbody radiation.

The Boltzmann factor $$\exp\left(-\frac{W}{k_BT}\right)$$ describes the probability with which certain energetic states are present!